尝试使用 mysqli 在 PHP 7.3 中启动事务时出现致命错误

Question

我想执行以下交易：

START TRANSACTION;
INSERT INTO
   `orders` (producten, totaalprijs, account_id) 
VALUES
   (
      "{\"601bced33b191\":{\"naam\":\"Sierkussens binnen\", \"alias\":\"sierkussens - binnen\", \"url\":\"https:\ / \ / printzelf.nl\ / new\ / product\ / sierkussens - binnen ? term = sierkussen\", \"afbeelding\":\"cms\ / images\ / producten\ / textiel_producten\ / Sierkussens\ / sierkussen_1.jpg\", \"aantal\":\"1\", \"hoogte\":null, \"breedte\":null, \"uploaden\":\"1\", \"specificaties\":{\"Formaat\":{\"waarde\":\"40 x 40 cm\"}, \"Materiaal\":{\"waarde\":\"Dekostof\"}, \"Samenstelling\":{\"waarde\":\"Hoes\"}, \"Kleur garen\":{\"waarde\":\"Zwart\"}, \"Kleur rits\":{\"waarde\":\"Zwart\"}, \"Ontwerp\":{\"waarde\":\"PRO ontwerpcontrole\"}}, \"prijs\":\"13.99\", \"totaalprijs\":13.99, \"canvas\":\"{\n \\"customer_id\\": \\"33\\", \n \\"order_id\\": \\"123\\", \n \\"quantity\\": \\"1\\", \n \\"rulers\\": null, \n \\"canvas_size\\": \\"0x0\\", \n \\"bleed\\": null, \n \\"safety_margin\\": null, \n \\"dpi\\": null, \n \\"multiple_layouts\\": null, \n \\"procheck\\": \\"y\\", \n \\"multiple_pages\\": \\"1\\", \n \\"product_name\\": \\"Sierkussens binnen\\", \n \\"thumbnail\\": \\"cms\\\ / images\\\ / producten\\\ / textiel_producten\\\ / Sierkussens\\\ / sierkussen_1.jpg\\"\n}\"}}", "19", "33" 
   )
;
UPDATE
   `orders` 
SET
   order_id = CONCAT(account_id, ".", id) 
WHERE
   id = LAST_INSERT_ID();
COMMIT;

所以我尝试开始这样的交易：

$conn->autocommit(FALSE);
$conn->query('
INSERT INTO orders
(
    producten,
    totaalprijs,
    account_id
)
VALUES
(
    "'.$conn->real_escape_string($encodedproductenarray).'",
    "'.$conn->real_escape_string(subtractbtw($subtotaalcheckoutafr)).'",
    "'.$conn->real_escape_string($_SESSION['user']['id']).'"
)');
$conn->query('
UPDATE orders SET order_id = CONCAT(account_id, ".", id)
WHERE id = LAST_INSERT_ID()');
$conn->commit();

但这给了我以下错误：

PHP Fatal error:  Uncaught Error: Call to undefined method Connection::autocommit()

我是不是做错了什么？

$conn 是我的连接 class，包含以下内容：

session_start();
class Connection {
    // Configure Database Vars
    private $host     = 'localhost';
    private $username = 'sdgsdg';
    private $password = 'sdgsdg!';
    private $db_name  = 'dsgsd';
    public $db;

    function __construct() {
        // Create connection
        $db = new mysqli($this->host, $this->username, $this->password, $this->db_name);
        // Check connection
        if ($db->connect_errno > 0) {
            die('Unable to connect to the database: '.$db->connect_error);
        }
        $this->db = $db;
    }

    public function query($query) {
        $db = $this->db;
        $this->db->query('SET NAMES utf8');
        if (!$result = $this->db->query($query)) {
            die('There was an error running the query ['.$db->error.']');
        } else {
            return $result;
        }
    }

    public function multi_query($query) {
        $db = $this->db;
        if (!$result = $this->db->multi_query($query)) {
            die('There was an error running the multi query ['.$db->error.']');
        } else {
            return $result;
        }
    }

    public function real_escape_string($value) {
        return $this->db->real_escape_string($value);
    }

    public function inserted_id() {
        return $this->db->insert_id;
    }
}

$conn = new Connection();

我也试过将整个交易添加到这样的查询中：

$insertorder = '
START TRANSACTION;
    INSERT INTO orders
    (
        producten,
        totaalprijs,
        account_id
    )
    VALUES
    (
        "'.$conn->real_escape_string($encodedproductenarray).'",
        "'.$conn->real_escape_string(subtractbtw($subtotaalcheckoutafr)).'",
        "'.$conn->real_escape_string($_SESSION['user']['id']).'"
    );
    UPDATE orders SET order_id = CONCAT(account_id, ".", id)
    WHERE id = LAST_INSERT_ID();
COMMIT;';
$insertordercon = $conn->query($insertorder);

但这给了我一个语法错误。

Answer 1

您的 class 中有一个 mysqli 对象。那有您要使用的所有方法。做这样的事情：

$conn->db->autocommit(FALSE);
$conn->query('
INSERT INTO orders
(
    producten,
    totaalprijs,
    account_id
)
VALUES
(
    "'.$conn->real_escape_string($encodedproductenarray).'",
    "'.$conn->real_escape_string(subtractbtw($subtotaalcheckoutafr)).'",
    "'.$conn->real_escape_string($_SESSION['user']['id']).'"
)');
$conn->query('
UPDATE orders SET order_id = CONCAT(account_id, ".", id)
WHERE id = LAST_INSERT_ID()');
$conn->db->commit();

或者，您可以在 class 中实现这些方法。您已经为 query 和 real_escape_string 这样做了，所以这应该很容易。只需插入 class:

public function autocommit($b) {
    $this->db->autocommit($b);
}

public function commit() {
   $this->db->commit();
}

...或类似的。

尝试使用 mysqli 在 PHP 7.3 中启动事务时出现致命错误

Fatal error when trying to start a transaction in PHP 7.3 with mysqli

php

sql

mysqli

transactions